$file = '../image.jpg'; $type = 'image/jpeg'; header('Content-Type:'.$type); header('Content-Length: ' . filesize($file)); readfile($file);

If you have the liberty to configure your webserver yourself, tools like mod_xsendfile (for Apache) are considerably better than reading and prin...

$file = '../image.jpg'; if (file_exists($file)) { $size = getimagesize($file); $fp = fopen($file, 'rb'); if ($size and $fp) {...

header('Content-type: image/jpeg'); readfile($image);

Try this: <?php header("Content-type: image/jpeg"); readfile("/path/to/image.jpg"); exit(0); ?>

For the next guy or gal hitting this problem, here's what worked for me: ob_start(); header('Content-Type: '.$mimetype); ob_end_clean(); $fp = fope...

(Expanding on the accepted answer ...) I needed to: log views of a jpg image and an animated gif , and, ensure that the images are neve...

You can use header to send the right Content-type : header('Content-Type: ' . $type); And readfile to output the content of the image : readfi...

$file = '../image.jpg'; $type = 'image/jpeg'; header('Content-Type:'.$type); header('Content-Length: ' . filesize($file)); $img = f...

Check out readfile() . The basic idea is you send the appropriate MIME type headers (using header()) then deliver the file contents using readfil...

Why not just reference the image directly then? If you are trying to hide the fact you are pulling an image from an external source, that external...

Just generate or read, then output the image using PHP. ...get image data from file or dynamically... header('Content-type: image/png'); //or what...

I've discovered problems if I didn't also include a 'Content-Length: ' header. The problems are crawler, proxy, and browser caching related. In wor...